Problem: Let $R$ be the region enclosed by the $x$ -axis, the line $x=4$, and the curve $y=\text{ln}(x)$. $y$ $x$ ${y=\text{ln}(x)}$ $ R$ $ 1$ $ 4$ A solid is generated by rotating $R$ about the $x$ -axis. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int^4_1 \ln(x)\cdot x\, dx$ (Choice B) B $\pi \int^{\text{ln}(4)}_1 [\ln(x)]^2\, dx$ (Choice C) C $\pi \int^4_1 [\ln(x)]^2\, dx$ (Choice D) D $\pi \int^{\text{ln}(4)}_1 \ln(x)\cdot x\, dx$
Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\text{ln}(x)}$ Each slice is a cylinder. Let the thickness of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\text{ln}(x)}$ $ 1$ $ 4$ $r$ The radius is equal to the distance between the curve $y=\text{ln}(x)$ and the $x$ -axis. In other words, for any $x$ -value, $r(x)=\text{ln}(x)}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(\text{ln}(x)}\right)^2 \end{aligned}$ The leftmost endpoint of $R$ is at $x=1$ and the rightmost endpoint is at $x=4$. So the interval of integration is $[1,4]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^4 \left[\pi\left(\text{ln}(x)\right)^2\right]dx \\\\ &=\pi\int_1^4 \left[\text{ln}(x)\right]^2\,dx \end{aligned}$